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Proof step 11 from proof get-finto (previous step)

m = m0 + 1, m ≤ # x0, m ≤ n0, n0 ≤ # x0, ∀ m, n. m < n ∧ m < # x0 ∧ n ≤ # x0 → (x0 to n)[m] = x0[m] ∨ x0 = [], ∀ m, x0, a. m < n0 ∧ m ≤ # x0 ∧ n0 ≤ 1 + # x0 ∧ (∀ m, n. m < n ∧ m < # x0 ∧ n ≤ # x0 → (x0 to n)[m] = x0[m] ∨ x0 = []) → (a ' + x0 to n0)[m] = (a ' + x0)[m] ⊦ (x0 to n0)[m0] = x0[m0], m = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
m0 < # x0, m0 < n0, n0 ≤ # x0, ∀ m, x0, a. m < n0 ∧ m ≤ # x0 ∧ n0 ≤ 1 + # x0 ∧ (∀ m, n. m < n ∧ m < # x0 ∧ n ≤ # x0 → (x0 to n)[m] = x0[m] ∨ x0 = []) → (a ' + x0 to n0)[m] = (a ' + x0)[m], ∀ m, n. m < n ∧ m < # x0 ∧ n ≤ # x0 → (x0 to n)[m] = x0[m] ⊦ (x0 to n0)[m0] = x0[m0], x0 = []

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)% ⊦ # x = 0 ↔ x = []; (s)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≠ 0; (s)